Understanding Delta Arguments in Limits: A Guide for Beginners

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In summary, the two problems the author is struggling with are that they can't seem to prove that a limit exists and that they don't understand how delta-N and delta-Epsilon arguments work.
  • #1
dcl
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I'm really stuck when it comes to proving things with delta-N or delta-Epsilon arguments.

I think my biggest problem is that I don't really see how they work or 'how' they prove the 'limit' or what have you.

Would anyone be able to show me how to do some of the following questions and somewhat explain what is going on..

Use delta-N arguments to prove that:
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^2 =0[/tex]

Use delta-epsilon arguements to prove that as x -> 3
[tex]5x \to 15[/tex]

Guess the limit and Use delta-epsilon arguments to prove your guess correct.
[tex]
\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}
[/tex]

Many thanks in advance. My notes don't make it clear how I'm supposed to do this :(
 
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  • #2
Well, first of all, you can't prove that
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^2 =0[/tex]
because it is not true. It should be obvious that that sequence has no limit. (If you had "-2" as the exponent instead of 2, then it would go to 0.)

The definition of "limit of a function" is "Given ε> 0, there exist a δ>0 such that if |x- x0|< δ then |f(x)- L|< ε. The "standard" proof of limits that you see in books starts from |f(x)-L|< &epsilong; and works backwards, calculating δ to show that it exists.

To show that [itex]\mathop {\lim }\limits_{n \to 3 } 5x =15[/itex] we need to get to |5x-15|< ε. Of course, |5x-15|= 5|x-3| so |5x-15|< ε is the same as |x-3|< &epsilon/5. We can take δ= &epsilon;/5.

[tex]\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}[/tex]
is considerably harder! First, because I know that [itex]\frac{1}{1+x^2}[/itex] is a continuous function, I would "guess" that the limit is [itex]\frac{1}{1+4^2}= \frac{1}{17}[/itex].
That means I want to arrive at [itex]|/frac{1}{1+x^2}-/frac{1}{17}|< \epsilon[/itex].

That is: [itex]|\frac{17- 1- x^2}{1+x^2}|= |\frac{x^2- 16}{1+x^2}|= |\frac{(x-4)(x+4)}{1+x^2}|= |x-4||\frac{x+4}{1+x^2}|< \epsilon[\itex].

The "x-4" term is exactly what we want. We have [itex]|x-4|< \epsilon\frac{x+4}{1+x^2}[\/tex] is close to 4, what must [itex]\frac{x+4}{1+x^2}[/itex] be close to?
 
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  • #3
HallsofIvy said:
[tex]\mathop {\lim }\limits_{x \to 4} \frac{1}{{1 + x^2 }}[/tex]
is considerably harder! First, because I know that [itex]\frac{1}{1+x^2}[/itex] is a continuous function, I would "guess" that the limit is [itex]\frac{1}{1+4^2}= \frac{1}{17}[/itex].
That means I want to arrive at [itex]|\frac{1}{1+x^2}-\frac{1}{17}|< \epsilon[/itex].

That is: [itex]|\frac{17- 1- x^2}{1+x^2}|= |\frac{x^2- 16}{1+x^2}|= |\frac{(x-4)(x+4)}{1+x^2}|= |x-4||\frac{x+4}{1+x^2}|< \epsilon[/itex].

The "x-4" term is exactly what we want. We have [itex]|x-4|< \epsilon\frac{1+x^2}{x+4}[/itex]; x is close to 4, what must [itex]\frac{1+x^{2}}{x+4}[/itex] be close to?

Just removing a few beauty spots..
 
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  • #4
I've probably learned more math and physics from reading posts and working problems that people have difficulty with, than sitting passively in class for hours.
 
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  • #5
Thanks for that, think I'm getting the hang of it...
Yeh, the first limit problem was a typo, it was meant to be:
[tex]\mathop {\lim }\limits_{n \to \infty } (n + 4)^-2 =0[/tex]
I don't suppose you could show me that, I 'think' I may have done it but I REALLY can't be sure :(
 
  • #6
You must show, for arbitrary [tex]\epsilon>0[/tex] that there exist an N so that for all n>N, [tex]\frac{1}{(n+4)^{2}}<\epsilon[/tex] :

1. [tex]\frac{1}{(n+4)^{2}}[/tex] is decreasing with n

This should be fairly easy to prove!
Hence, if you are able to find an N which satisfy [tex]\frac{1}{(N+4)^{2}}<\epsilon[/tex] (for a given [tex]\epsilon>0[/tex]), you have also shown it for n>N

2. We want to find N so that:
[tex]\frac{1}{(N+4)^{2}}<\epsilon[/tex]

This inequality is equivelent to:
[tex]\frac{1}{\epsilon}<(N+4)^{2}[/tex]

Hence, we find the requirement on N:
[tex]N>\sqrt{\frac{1}{\epsilon}}-4[/tex]
 
  • #7
Nice. I'm suprised it turned out to be that simple...
 

What is a delta argument?

A delta argument is a type of scientific argument that involves making a comparison or contrast between two different states or conditions. It is often used to explain changes or differences in data or observations.

Why are delta arguments important?

Delta arguments are important because they allow scientists to examine and understand changes or differences in data over time. This can provide valuable insights into the natural world and help us make predictions about future events or phenomena.

How do you construct a delta argument?

A delta argument typically involves identifying and describing the two states or conditions being compared, providing evidence for the changes or differences between them, and drawing conclusions based on the evidence. It is important to use clear and logical reasoning in constructing a delta argument.

What types of data are often used in delta arguments?

Delta arguments can be based on many different types of data, including quantitative data (such as measurements or numerical values) and qualitative data (such as observations or descriptions). The type of data used will depend on the specific research question or topic being studied.

Can delta arguments be used in different scientific disciplines?

Yes, delta arguments can be used in a variety of scientific disciplines, including biology, chemistry, physics, and environmental science. They are a versatile tool for analyzing and interpreting data in many different fields of study.

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